\(\Leftrightarrow\left\{{}\begin{matrix}x^3+3xy^2=-49\\3x^2-24xy+3y^2=24y-51x\end{matrix}\right.\)

Cộng vế:

\(x^3+3x^2+3y^2\left(x+1\right)-24y\left(x+1\right)+51x+49=0\)

\(\Leftrightarrow\left(x+1\right)^3+3y^2\left(x+1\right)-24y\left(x+1\right)+48\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)^3+3\left(x+1\right)\left(y-4\right)^2=0\)

\(\Leftrightarrow\left(x+1\right)\left[\left(x+1\right)^2+3\left(y-2\right)^2\right]=0\)

c) \(3x+3y-x^2-2xy-y^2=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)d) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)

\(=\left(x+y\right)\left(x+y+1\right)\left(x+y-1\right)\)

\(c,=3\left(x+y\right)-\left(x+y\right)^2=\left(3-x-y\right)\left(x+y\right)\\ d,=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)

c) \(=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)

d) \(\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)

http://diendantoanhoc.net/topic/151610-leftbeginmatrix-x33xy2-49-x2-8xyy28y-17x2-endmatrixright/

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`a, x^3 + y^3 + x + y`

`= (x+y)(x^2-xy+y^2)+x+y`

`= (x+y)(x^2-xy+y^2+1)`

`b, x^3 - y^3 + x -y`

`= (x-y)(x^2+xy+y^2)+x-y`

`= (x-y)(x^2+xy+y^2+1)`

`c, (x-y)^3 + (x+y)^3`

`= (x-y+x+y)(x^2-2xy+y^2 - x^2 + y^2 + x^2 + 2xy + y^2)`

`= (2x)(x^2 + 3y^2)`

`d, x^3 - 3x^2y + 3xy^2 - y^3 + y^2 - x^2`

`= (x-y)^3 + (y-x)(x+y)`

`=(x-y)(x^2+2xy+y^2-x-y)`

a: =(x+y)(x^2-xy+y^2)+(x+y)

=(x+y)(x^2-xy+y^2+1)

b: =(x-y)(x^2+xy+y^2)+(x-y)

=(x-y)(x^2+xy+y^2+1)

c: =x^3-3x^2y+3xy^2-y^3+x^3+3x^2y+3xy^2-y^3

=2x^3+6xy^2

d: =(x-y)^3+(y-x)(y+x)

=(x-y)[(x-y)^2-(x+y)]

nhân pt (2) vs 3 sau đó cộng pt (1) vs (2) ta đc

\(\left\{{}\begin{matrix}x^3+3xy^2=-46\\x^3+3xy^2+3x^2-24xy+3y^2=24y-51x-46\end{matrix}\right.\)

bây h ta chú ý tới pt dưới

\(x^3+3xy^2+3x^2-24xy+3y^2-24y+51x+46=0\)

\(\left(x+1\right)\left(x^2+2x+3y^2-24y+49\right)=0\)

\(\left(x+1\right)\left[\left(x+1\right)^2+3\left(y-4\right)^2\right]=0\)

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-1\\x^3+3xy^2=-49\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=4\end{matrix}\right.\end{matrix}\right.\rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-1\\y=-4\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=4\end{matrix}\right.\end{matrix}\right.\)

vậy hệ có 2 nghiệm

Đây nè bạn.

a) Ta có: \(M=x^2-2xy+y^2-10x+10y\)

\(=\left(x-y\right)^2-10\left(x-y\right)\)

\(=9^2-10\cdot9=-9\)

\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)

\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11

e: Ta có: \(x^2-6x+y^2+4y+2=0\)

\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Dấu '=' xảy ra khi x=3 và y=-2